Answer: The mass of bromine reacted is 160.6 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of aluminium = 18.1 g
Molar mass of aluminium = 27 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol[/tex]
The chemical equation for the reaction of aluminium and bromide follows:
[tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]
By Stoichiometry of the reaction:
2 moles of aluminium reacts with 3 moles of bromine gas
So, 0.670 moles of aluminium will react with = [tex]\frac{3}{2}\times 0.670=1.005mol[/tex] of bromine gas.
Now, calculating the mass of bromine gas, we use equation 1:
Moles of bromine gas = 1.005 moles
Molar mass of bromine gas = 159.81 g/mol
Putting values in equation 1, we get:
[tex]1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g[/tex]
Hence, the mass of bromine reacted is 160.6 grams.
Final answer:
To find out how many grams of bromine react with 18.1g of aluminum, we first determine the mass ratio from a previous reaction and then use that to calculate the mass of bromine. The proportional relationship indicates that approximately 159.8 grams of bromine would react with 18.1 grams of aluminum.
Explanation:
The student asks how many grams of bromine react with 18.1g of aluminum. The chemical reaction is similar in behavior to the reaction of aluminum with chlorine where aluminum bromide is formed. Given a previous reaction scale of 27g aluminum yielding 266.7g aluminum bromide, we use stoichiometry to find the grams of bromine reacting with 18.1g aluminum.
First, determine the molar mass of aluminum (Al) and aluminum bromide (AlBr3). Aluminum has a molar mass of approximately 27g/mol and aluminum bromide has a molar mass of roughly 267g/mol based on the atomic weights of aluminum (approximately 27) and bromine (approximately 80 per bromine atom, with three bromine atoms per molecule). Find the mole ratio of Al to AlBr3 from the balanced chemical equation, which is 2:2 (or 1:1 for simplicity). From the initial reaction scale, you can calculate the ratio of aluminum to bromine involved in the reaction.
Next, set up a proportional relationship:
27g Al : 266.7g AlBr3 = 18.1g Al : x g AlBr3
To solve for x (the mass of aluminum bromide produced from 18.1g of aluminum), cross-multiply and divide:
27g Al * x g AlBr3 = 18.1g Al * 266.7g AlBr3
x = (18.1g Al * 266.7g AlBr3) / 27g Al
x ≈ 177.7g AlBr3
Finally, determine the mass of bromine involved. Since 27g of Al produces 266.7g of AlBr3, the mass of bromine can be calculated by subtracting the mass of aluminum from the total mass of the product:
266.7g AlBr3 - 27g Al = 239.7g Br2
Now, calculate the mass of bromine that would react with 18.1g of Al using the same proportion:
27g Al : 239.7g Br2 = 18.1g Al : y g Br2
Cross-multiply and solve for y:
y = (18.1g Al * 239.7g Br2) / 27g Al
y ≈ 159.8g Br2
Therefore, approximately 159.8 grams of bromine would react with 18.1 grams of aluminum.
Earth's surface area is 5.10 × 108 km2, and its crust has a mean thickness of 35 km and mean density of 2.8 g/cm3. The two most abundant elements in the crust are oxygen (4.55 × 105 g/t) and silicon (2.72 × 105 g/t), and the two rarest non-radioactive elements are ruthenium and rhodium, each with an abundance of 1 × 10−4 g/t. What is the total mass of each of these elements in Earth's crust? 1 metric ton (t) = 1,000 kg.
The mass of oxygen, silicon, ruthenium, and rhodium in the earth's crust is 2.274 × 1031 g, 1.360 × 1031 g, 4.998 × 1021 g, and 4.998 × 1021 g respectively.
Explanation:To find the total mass of each of these elements in Earth's crust, we must first calculate the total volume of the crust. This is done by multiplying the surface area of the Earth by the mean thickness of the crust: 5.10 × 108 km2 * 35 km = 1.785 × 1010 km3.
Next, we convert this volume into cm3, because the density is given in g/cm3: 1.785 × 1010 km3 * (105)3 = 1.785 × 1025 cm3.
Then, we can calculate the total mass of the crust by multiplying the volume by the density: 1.785 × 1025 cm3 * 2.8 g/cm3 = 4.998 × 1025 g.
From there, we can calculate the mass of each element by multiplying the total mass by the abundance of each element. For oxygen and silicon, this would be: 4.998 × 1025 g * 4.55 × 105 g/t = 2.274 × 1031 g for oxygen and 4.998 × 1025 g * 2.72 × 105 g/t = 1.360 × 1031 g for silicon. For the rare elements ruthenium and rhodium, this would be: 4.998 × 1025 g * 1 × 10-4 g/t = 4.998 × 1021 g for both elements.
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Observe the following balanced chemical reaction:
2 Fe + 3 Cl2 → 2 FeCl3
PLEASE HELP ASAP
Which of the following describes the mole ratio between Fe and Cl2?
2 mol Fe: 2 mol Cl2
1 mole Fe: 2 mole Cl2
2 mole Fe: 6 mol Cl2
2 mole Fe : 3 mol Cl2
Answer:
The answer to your question is: the last option ( 2 mole Fe : 3 mol Cl2)
Explanation:
The mole ratio in a chemical reaction is expressed by the integers before each element, for example if we have
2H2SO4 + 3NaOH,
The mole ratio between, H2SO4 and NaOH is 2H2SO4 : 3 NaOH
In your question, the question is the ratio between Fe and Cl2, so we look at their coefficients and then:
2 mole Fe : 3 mol Cl2
Which compounds are most likely to have covalent bonds?
Select all that apply.
Hint: Review your periodic table.
NaF
MgO
H2O
CO
Answer:
The answer to your question is; H2O and CO.
Explanation:
Ionic bonding is when a metal donates its valence electrons to a nonmetal.
Covalent bonding is when two nonmetals share electrons to reach noble gases' configuration.
NaF here we have a metal and a nonmetal, then the bonding is ionic.
MgO here, there is also a metal and a nonmetal, the bonding is ionic.
H2O here, there are two nonmetals, so this is a covalent bonding.
CO here, there are two nonmetals, the bonding is covalent.
Final answer:
H₂O and CO are the compounds most likely to have covalent bonds because they are composed of non-metallic elements. NaF and MgO are ionic compounds as they contain a metal and a non-metal.
Explanation:
Compounds that are most likely to have covalent bonds are usually formed between non-metallic elements. In the choices given, compounds H₂O (water) and CO (carbon monoxide) are most likely to have covalent bonds because they consist of non-metallic elements bonded together.
Let's evaluate the options further:
Sodium fluoride (NaF) - Ionic compound due to the metal (Na) and non-metal (F) combination.Magnesium oxide (MgO) - Ionic compound, as it consists of a metal (Mg) reacting with a non-metal (O).Water (H₂O) - Covalent compound, made from non-metals hydrogen and oxygen.Carbon monoxide (CO) - Covalent compound, with carbon and oxygen being non-metals.In an air-conditioned room at 19.0 ∘C, a spherical balloon had the diameter of 50.0 cm. When taken outside on a hot summer day, the balloon expanded to 51.0 cm in diameter. What was the temperature outside in degrees Celsius?
Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.
The temperature outside on the hot summer day was approximately 37.18 °C.
Charles's law, which states that for constant pressure and volume of a gas, the volume of a gas is proportional to its absolute temperature (measured in Kelvin), can be used to resolve this:
Calculating the temperature outside ([tex]\rm T_2[/tex]) in Kelvin using the information provided:
Initial temperature ([tex]\rm T_1[/tex]) = 19.0 °C + 273.15 K = 292.15 K
Initial diameter ([tex]\rm d_1[/tex]) = 50.0 cm
Final diameter ([tex]\rm d_2[/tex]) = 51.0 cm
[tex]\rm (T_1 / T_2) = (d_1^3 / d_2^3)[/tex]
[tex]\rm (292.15 K / T_2) = (50.0^3 / 51.0^3)[/tex]
Calculating the right-hand side of the equation:
[tex](50.0^3 / 51.0^3)[/tex] ≈ 0.94149
Solving for [tex]\rm T_2[/tex]:
[tex]\rm T_2[/tex] = 292.15 K / 0.94149 ≈ 310.33 K
Converting [tex]\rm T_2[/tex] back to Celsius:
Temperature outside (in Celsius) = [tex]\rm T_2[/tex] - 273.15
Temperature outside (in Celsius) = 310.33 K - 273.15 K ≈ 37.18 °C
Therefore, the temperature outside on the hot summer day was approximately 37.18 °C.
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The problem is based on Charles's Law which states the volume of a gas is directly proportional to its temperature at constant pressure. The initial and final conditions of the balloon are used to calculate the outside temperature to be approximately 38.35°C.
Explanation:To solve this problem, we need to use the concepts of gas laws, specifically Charles's Law. Charles's Law states that volume of a gas is directly proportional to its temperature if pressure and the amount of gas remain constant. In this case, we assume that the pressure inside the balloon remains constant because it is exposed to constant external atmospheric pressure both inside the air-conditioned room and outside.
Let's express the initial and final conditions as follows:
Initial volume (V1) = 4/3*pi*(diameter/2)^3 = 4/3*pi*(50/2)^3 = 65450 cm^3Final volume (V2) = 4/3*pi*(diameter/2)^3 = 4/3*pi*(51/2)^3 = 69813 cm^3Initial temperature (T1) = 19.0 °C = 19.0 + 273.15 = 292.15 KFinal temperature (T2) = ? KAccording to Charles's Law, V1/T1=V2/T2. So, we can find the final temperature as T2 = V2*(T1/V1). Substituting our values, we find T2 = 69813*(292.15/65450) = 311.5 K. Finally, convert it back to Celsius by subtracting 273.15: T2 = 311.5 - 273.15 = 38.35°C. So, it was approximately 38.35°C outside.
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can someone please help me with these questions. image attached
In a laboratory experiment, a fermenting aqueous solution of glucose and yeast produces carbon dioxide gas and ethanol. The solution was heated by burning natural gas in a Bunsen burner to distill the ethanol that formed in the flask. During the distillation, the ethanol evaporated and then condensed in the receiving flask. The flame of the burner was kept too close to the bottom of the flask and some of the glucose decomposed into a black carbon deposit on the inside of the flask. During this experiment the following changes occurred. Which of these changes involved a physical change and not a chemical change?
1. evaporation of ethanol
2. condensation of ethanol
3. formation of a carbon deposit inside the flask
4. formation of carbon dioxide gas from glucose
5. burning of natural gas
6. formation of ethanol from glucose by yeast
Answer:
1. evaporation of ethanol
2. condensation of ethanol
Explanation:
The first and second process imply a change of phase, which is essentially a physical change because the matter change from liquid to gas and from gas to liquid respectively and it can ve reversed.
The options 3 to 6 are related to a chemical reaction because the outcomes of the process are different compounds than the ones we had at the beginning. The main reaction that implies process 3 to 6 is represented by:
C6H12O6 + impurities → 2 C2H5OH + 2 CO+ C
For option 5. burning of natural gas , we have:
Natural gas(Mainly a mix of CH4 and C2H10) +O2→COn+H2O
Therefore, justo 1 and 2 are a physical change
The evaporation and condensation of ethanol are physical changes as these do not alter the substance's chemical composition. The formation of carbon deposit, carbon dioxide gas, the burning of natural gas, and ethanol formation all involve chemical reactions and are thus considered chemical changes.
Explanation:In the experiment described, both physical and chemical changes occur. Physical changes involve a change in physical state without altering the substance's chemical composition. Therefore, the evaporation of ethanol (#1), and the condensation of ethanol (#2) are physical changes. The substance remains ethanol in both cases; it merely changes from liquid to gas or from gas to liquid.
Chemical changes, on the other hand, involve a chemical reaction where a new substance is formed. Therefore, the formation of a carbon deposit inside the flask from decomposition of glucose (#3), the formation of carbon dioxide gas from glucose (#4), the burning of natural gas (#5), and the formation of ethanol from glucose by yeast (#6) are examples of chemical changes.
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The density of a solution of sulfuric acid is 1.29 g/cm3 and it is 38.1% acid by mass. What volume of the sulfuric acid solution is needed to supply 163 g of sulfuric acid? 1 cm3 = 1 mL 1. 428 g 2. 80.1 mL 3. 552 mL 4. 252 mL 5. 48.1 mL 6. 8010 mL 7. 332 mL 8. 0.00397 mL
Answer : The correct option is, (7) 332 mL
Explanation : Given,
Density of solution of sulfuric acid = [tex]1.29g/cm^3=1.29g/mL[/tex]
38.1 % acid by mass that means 38.1 grams of sulfuric acid present in 100 grams of solution of sulfuric acid.
Now we have to calculate the mass of solution of sulfuric acid.
As, 38.1 grams of sulfuric acid present in 100 grams of solution of sulfuric acid
So, 163 grams of sulfuric acid present in [tex]\frac{163}{38.1}\times 100=428.95[/tex] grams of solution of sulfuric acid
Now we have to calculate the volume of sulfuric acid solution.
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]1.29g/mL=\frac{428.95g}{Volume}[/tex]
[tex]Volume=332mL[/tex]
Therefore, the volume of sulfuric acid solution needed is 332 mL.
Which of the following is NOT a derived unit?
a) [tex]cm^{3}[/tex] b) g/mL c)kg
Answer:
c) kg
Explanation:
Kilograms stands alone. It has to be hooked up to another unit for it to be a derived unit.
I am joyous to assist you anytime.
Answer:
The answer to your question is c) kg
Explanation:
Derived units is when to units are combined to get a new one, For example, combining length and length gives lenght² and then we measured area, this is a derived unit.
a) cm³ this answer is wrong because we are combining length x length x length so we get the units of volume, this is a derived unit.
b) g/mL this answer is wrong because we are combining mass and volume so we can measure density, this units are derived.
c)kg this answer is correct because it measures mass and only that, then it's a fundamental unit.
Chemical digestion of carbohydrates and lipids begins in the ____________, while chemical digestion of proteins begins in the _____________.
Answer:
mouth, stomach
Explanation:
The carbohydrates start the digestion in the mouth, with the enzymes dissolved in the saliva, and during the rest of the digestive system it continues until the intestine.
Most of the digestion of lipids is in the intestine, with the lipid enzymes that come from the bile. But it starts in the mouth with the absorption of the small ones.
The chemical digestion of the proteins begins in the stomach because the peptide enzyme it turns active with the acidic environment of the stomach.
How many liters of air are in:
A) a room that measures 10.0 ft × 11.0 ft and has an 8.00 ft ceiling?
B) a room that measures 11.0 ft x 11.0 ft and has a 10.0 ft ceiling?
1 in. = 2.54 cm (exactly); 1 L = 103cm3
Calculate the volume of air in two different rooms based on their dimensions in feet. The volume of air in each room is converted to liters using the given conversion factor. The first room has approximately 24,900 liters of air, and the second room has around 34,300 liters of air.
The volume of air in a room can be calculated using the formula:
Volume = Length x Width x Height
A) For a room measuring 10.0 ft x 11.0 ft x 8.00 ft:
Volume = 10.0 ft x 11.0 ft x 8.00 ft = 880 ft³ = 24,900 L
B) For a room measuring 11.0 ft x 11.0 ft x 10.0 ft:
Volume = 11.0 ft x 11.0 ft x 10.0 ft = 1,210 ft³ = 34,300 L
In a chemical reaction, substrate molecule A is broken down to form one molecule of product B and one molecule of product C. The equilibrium constant, K, for this reaction is 0.5. If we start with a mixture containing only substrate A at a concentration of 1 M, what will be the concentration of A when the reaction reaches equilibrium?
Answer:
Concentration of A at equilibrium = 1 - 0.5 = 0.5 M
Explanation:
[tex]A \leftrightharpoons B + C[/tex]
Equilibrium constant = 0.5
Initial concentration of A = 1 M
[tex]A \leftrightharpoons B + C[/tex]
Initial 1 0 0
At equi. 1-x x x
Equilibrium constant = [tex]\frac{[B][C]}{[A]}[/tex]
[tex]0.5 = \frac{x \times x}{1-x} \\0.5(1-x) = x^2\\0.5 -0.5x = x^2\\x^2+0.5x - 0.5 = 0[/tex]
on solving,
x = 0.5 M
Concentration of A at equilibrium = 1 - 0.5 = 0.5 M
Final answer:
Use the equilibrium constant and an ICE table to set up the equation Keq = [B][C]/[A] and solve for x to determine the equilibrium concentration of A after adding more of the substrate to the mixture.
Explanation:
To calculate the concentration of A when the reaction reaches equilibrium after adding 0.50 moles of A, we use an ICE table for the equilibrium process and the equilibrium constant given as Keq = 0.5. Initially, we have 1.0 M of A without B. Upon addition, we have 1.50 M of A. If we let x be the change in concentration of A as it reacts to form B and C, at equilibrium, [A] = 1.50 - x, [B] = x, and [C] = x because the mole ratio of A:B:C is 1:1:1. Setting up the equilibrium expression as Keq = [B][C]/[A], and substituting the terms with the expressions in terms of x, we get 0.5 = x²/(1.50 - x). We must solve this quadratic equation to find the value of x and, consequently, the equilibrium concentration of A.
Equilibrium constant = ([B][C])/([A])
0.5 = (x * x)/(1-x) \n0.5(1-x) = x^2\n0.5 -0.5x = x^2\nx^2+0.5x - 0.5 = 0
on solving,
x = 0.5 M
Concentration of A at equilibrium = 1 - 0.5 = 0.5 M
A 0.4987-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.9267 g of CO₂ and 0.1897 g of H₂O. What is the empirical formula of the compound?
Answer: The empirical formula for the given compound is [tex]C_3H_3O_2[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=0.9267g[/tex]
Mass of [tex]H_2O=0.1897g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 0.9267 g of carbon dioxide, [tex]\frac{12}{44}\times 0.9267=0.2527g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 0.1897 g of water, [tex]\frac{2}{18}\times 0.1897=0.021g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (0.4987) - (0.2527 + 0.021) = 0.225 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.2527g}{12g/mole}=0.021moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.021g}{1g/mole}=0.021moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.225g}{16g/mole}=0.014moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.
For Carbon = [tex]\frac{0.021}{0.014}=1.5[/tex]
For Hydrogen = [tex]\frac{0.021}{0.014}=1.5[/tex]
For Oxygen = [tex]\frac{0.014}{0.014}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1.5 : 1.5 : 1
To make in whole number we multiply the ratio by 2, we get:
The ratio of C : H : O = 3 : 3 : 2
The empirical formula for the given compound is [tex]C_3H_3O_2[/tex]
Thus, the empirical formula for the given compound is [tex]C_3H_3O_2[/tex]
PLEASE HELP WILL MARK BRAINLIEST
Claire walks into the kitchen after hearing a loud noise. She noticed a spilled glass of milk on the floor and says that the cat must
have been on the counter and knocked the glass over.
Claire’s statement can be classified as a
A) observation
B) prediction
C) inference
D) hypothesis
2)
Jack walks outside in the afternoon and noticed the clouds are very dark. He states that it looks like it will rain very soon.
His statement can be classified as
A) prediction
B) observation
C) inference
D) hypothesis
Answer:
1 is b and 2 is d
Explanation:
because you do not know that the cat had done that you was just thinking it was the cat
Answer:
The answer is
Explanation:
Definitions
Observation: is when something catches our attention
Prediction: is a statement about a future event, based upon experience and knowledge
Inference: is a conclusion based on evidence and reasoning.
Hypothesis: is a proposed explanation for a fenomenon
Case I:
It isn't an observation
It isn't a prediction because it happened not will happen
I think is an inference
It isn't a hypothesis, a hypothesis has an specific structure If ... then ...
Case II
I think is a prediction, something that will happen
It isn't an observation he did more than that
It isn't an inference
Hypothesis has an structure.
Methane and ethane are both made up of carbon and hydrogen. In methane, there are 12.0 g of carbon for every 4.00 g of hydrogen, a ratio of 3:1 by mass. In ethane, there are 24.0 g of carbon for every 6.00 g of hydrogen, a ratio of 4:1 by mass. This is an illustration of the law of __________.
Answer:
The answer is: Law of multiple proportions
Explanation:
The law of multiple proportions is a law of chemical combination given by Dalton in 1803.
According to this law, if more than one chemical compound is formed by combining two elements, then the mass of an element that combines with the fixed mass of other element is represented in the form of small whole number ratio.
Therefore, is an illustration of the law of the law of multiple proportions.
Methane and ethane are both made up of carbon and hydrogen. In methane, there are 12.0 g of carbon for every 4.00 g of hydrogen, a ratio of 3:1 by mass. In ethane, there are 24.0 g of carbon for every 6.00 g of hydrogen, a ratio of 4:1 by mass. This is an illustration of the law of multiple proportions.
The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.
In the given example, methane [tex]CH_3[/tex] and ethane [tex]C_2H_6[/tex] are compounds formed by carbon (C) and hydrogen (H). The mass ratios of carbon to hydrogen in these compounds are as follows:
For methane :
- Mass of carbon (C) = 12.0 g
- Mass of hydrogen (H) = 4.00 g
- Mass ratio of C to H = 12.0 g / 4.00 g = 3:1
For ethane
- Mass of carbon (C) = 24.0 g
- Mass of hydrogen (H) = 6.00 g
- Mass ratio of C to H = 24.0 g / 6.00 g = 4:1
The ratios of the masses of carbon that combine with a fixed mass of hydrogen (12.0 g of carbon for methane and 24.0 g of carbon for ethane) are in a simple whole number ratio of 1:2 (or 3:6 if we consider the hydrogen masses as well). This exemplifies the law of multiple proportions, as the mass of carbon combining with a fixed mass of hydrogen increases in a small whole number ratio from one compound to the other.
Which of the following are examples of homogeneous mixtures? Cheddar cheese Iced tea with no ice Chicken noodle soup Bread pudding Which of the following are examples of heterogeneous mixtures? Vanilla ice cream Antifreeze Rocky road ice cream Trail mix
Cheddar cheese, iced tea with no ice, chicken noodle soup, vanilla ice cream, antifreeze, rocky road ice cream, and trail mix are examples of homogeneous and heterogeneous mixtures.
Explanation:
Homogeneous mixtures:
Cheddar cheeseIced tea with no iceChicken noodle soupBread puddingHeterogeneous mixtures:
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Homogeneous mixtures are uniform throughout; examples include iced tea with no ice and antifreeze. Heterogeneous mixtures aren't uniform and individual components can be seen; examples include cheddar cheese, chicken noodle soup, bread pudding, vanilla ice cream, rocky road ice cream, and trail mix.
Explanation:The question pertains to the classification of various items as either homogeneous or heterogeneous mixtures. A homogeneous mixture is a combination of substances with a composition that is uniform throughout. In such a mixture, you cannot see the individual components, and there's no variation from point to point. On the other hand, a heterogeneous mixture is a collection of substances whose composition can vary from point to point and where the individual components can be seen.
From the provided list, the examples of homogeneous mixtures are Iced tea with no ice and Antifreeze because their compositions are the same throughout. Meanwhile, Cheddar Cheese, Chicken noodle soup, Bread pudding, Vanilla ice cream, Rocky road ice cream, and Trail mix could be considered examples of heterogeneous mixtures as their compositions are not uniform and the different components can be seen distinctly.
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Which of the following statements is true?a. Acids and bases cannot mix together. b. Acids and bases will neutralize each other. c. Acids, but not bases, can change the pH of a solution. d. Acids donate hydroxide ions (OH–); bases donate hydrogen ions (H+).
Answer:
The answer is letter b.
Explanation:
a. Acids and bases cannot mix together. This is wrong, in fact there is a reaction called neutralization in which the reactants are an acid and a base.
b. Acids and bases will neutralize each other. This is true, when acids and bases react, the products are a salt and water and the reaction is called neutralization reaction.
c. Acids, but not bases, can change the pH of a solution. It's false, both acids and bases change the pH of a solution.
d. Acids donate hydroxide ions (OH–); bases donate hydrogen ions (H+). It's wrong, acids donate H+ and bases donate OH-.
The correct statement here is Acids and bases will neutralize each other.
An acid can be explained as substance which can gives protons or one that can accepts a pair of valence electrons to form a bond A base can be regarded as substance which can take protons or substance which can give a pair of valence electrons when forming bond. Acid and base cannot be be described as one that mix together, though they can undergo what is called neautralization reaction.Neutralization occurs when their is a reaction between acid and base whereby salt and water is been produced.The statement in the questions which says that Acid but not base can bring change to PH of solution is not true this is because both the acid as well as the base brings change to the PH of solution.An acid will donate H+ in a solution and base will give OH- in solution. I.e hydrogen ion and Hydroxide ion respectivelyLearn more at https://brainly.com/question/16239681?
If you did not know the identity of the metal in your jar, what experiment might you have done to prove that the metal was in fact the element pure zinc?
Answer:
Explanation:
I could have set up an experiment to determine the density of the metal in the jar. When I derive the density of the metal, I will compare it with that of every other metal to see if it properly fits any.
Density is the mass per unit volume of substance. It is the amount of substace contained per unit volume.
To find the density of the metal, the mass of the metal is obtained by directly weighing it. The volume is found by immersing the metal in water as it will sink. The volume of the water displaced is the volume of the metal. A measuring cylinder or an overflow is used to determine the volume of the liquid displaced.
Then:
Density = [tex]\frac{mass}{volume}[/tex]
Final answer:
To confirm the identity of the metal as pure zinc, perform an experiment with hydrochloric acid, observe reactions, record properties, and differentiate based on reactivity and density.
Explanation:
Experiment:
Place a few drops of zinc metal in a test tube and cover with dilute hydrochloric acid.
Heat the test tube containing the mixture to observe reactions and record your observations.
Wait for the product to cool, break the test tube, and examine the product to verify if it matches properties of pure zinc.
Observations:Zinc's reactivity with hydrochloric acid, its density compared to water, and its behavior in displacement reactions can be used to differentiate and confirm its identity as pure zinc.
Chemical Reactions:When zinc metal reacts with sulfuric acid, it produces zinc sulfate and hydrogen gas; whereas zinc oxide reacts with sulfuric acid to form zinc sulfate only.
If 10.6 kg of Al2O3(s),10.6 kg of Al2O3(s), 51.4 kg of NaOH(l),51.4 kg of NaOH(l), and 51.4 kg of HF(g)51.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?
Answer:
43.668 kg
Explanation:
First we set the equation:
[tex]Al_{2}O_{3} + 6NaOH + 12HF \longrightarrow 2Na_{3}AlF_{6}+9H_{2}O[/tex]
Now, we need to now the kmoles for each reactant:
[tex]M_{Al_{2}O_{3}}=101.96kg/kmol\\M_{NaOH}=40kg/kmol\\M_{HF}=20.01kg/kmol\\n_{Al_{2}O_{3}}=0.104kmol\\n_{NaOH}=1.285kmol\\n_{HF}=2.57kmol[/tex]
With this, we can see that the limit reactant is the aluminum oxide, so, with the equation for the reaction we know that 1 kmol of aluminum oxide, produces 2 kmol of cryolite, so we set a rule of three and see that 0.208 kmoles of cryolite are produced, the we proceed to calculate the mass:
[tex]M_{Na_{3}AlF_{6}}=209.94kg/kmol\\n_{Na_{3}AlF_{6}}=0.208kmol\\m_{Na_{3}AlF_{6}}=43.668kg[/tex]
The activated complex is located at point:
Enthalpy
Reaction Coordinate
Answer:
The answer to your question is letter B
Explanation:
Activated complex is an intermediate process between the reactants and the products which has a higher energy than both (reactants and products).
In the graph we can see that the higher point is letter B, that is the one of the activated complex. The energy that must be supply to the systed to proceed is called activated energy.
Which of the following, when dissolved, will cause water to reach the highest temperature before boiling?
A. Epsom Salt (MgSO4)
B. Potassium sulfide (K2S)
C. Sodium Chloride (NaCl)
D. Sugar (C6H12O6)
Answer:
B
Explanation:
The boiling point of water increases when a solute is dissolved in the water. The amount by which the boiling point is raised depends on the number of solute particles in the water.
For equal molar quantities of Epsom salt, potassium sulfide, sodium chloride, and sugar are, potassium sulfide yields the most particles.
K₂S dissociates as follows:
K₂S ⇒ 2K⁺ + S²⁻
One mole of K₂S yields 3 moles of particles.
Epsom salt (MgSO₄) yields 2 moles of particles for every mole:
MgSO₄ ⇒ Mg²⁺ + SO₄²⁻
Sodium chloride (NaCl) also yields 2 moles of particles for every mole:
NaCl ⇒ Na⁺ + Cl⁻
Sugar is not an electrolyte and does not dissociate, so 1 mole of sugar produces 1 mole of solute particles.
Answer:
Potassium sulfide (K2S)
Explanation:
Answer via Founder's Education/Educere
Stacy runs a paper chromatography experiment using the same ink, but two different solvents. Using solvent A results in the ink separating into three bands, but solvent B does not carry the ink away from the original spot.
Which of the following could explain the results?
a) Solvent A was not a good mobile phase because it could not dissolve the ink.
b) Solvent B was not a good mobile phase because it could not dissolve the ink.
c)The ink was a pure substance and dissolved in both solvents.
d) The ink was a mixture, but only in solvent A.
Answer:
b) Solvent B was not a good mobile phase because it could not dissolve the ink.
Explanation:
The base for chromatography is that the mixture to analyze is soluble in the solvent, if this happen, there will be separation. The only thing that may affect the separation, is that the mixture is not soluble in the solvent.
Answer: Option B
Explanation:
The paper chromatography can be defined as the technique in which there are two phase. One is stationary phase that lies in the cellulose fiber of the paper.
The mobile phase is the solvent that moves on the stationary phase. This is the usually solvent.
The interaction between the mobile phase and stationary phase must be good so as to carry the stationary phase along with the mobile phase.
If the stationary phase is not carried along with the mobile phase then the interaction between the stationary phase and mobile phase is not strong.
Methane and chlorine react to form four products: CH3Cl, CH2Cl2, CHCl3, and CCl4. At a particular temperature and pressure, 38.4 g of CH4 was allowed to react with excess Cl2 and gave 9.2 g CH3Cl, 47.1 g CH2Cl2, and 109 g CHCl3. All the CH4 reacted. (Note: The hydrogen that is displaced from the carbon also combines with Cl2 to form HCl.)How many grams of CCL4 were formed?How many grams of Cl2 reacted with the CH4?
Answer:
How many grams of CCL4 were formed? 116.9 g
How many grams of Cl2 reacted with the CH4? 243.8 g
Explanation:
First we need to know the molar mass for every element or compound in the reaction:
[tex]M_{CH_{4}}=16 g/mol\\M_{CH_{3}Cl}=50.49g/mol\\M_{CH_{2}Cl_{2}}=84.93g/mol\\M_{CHCl_{3}}=119.38g/mol\\M_{CCl_{4}}=153.82g/mol[/tex]
Now we proceed to calculate the amount of moles produced, per product:
[tex]n_{CH_{4}}=2.4\\n_{CH_{3}Cl}=0.18\\n_{CH_{2}Cl_{2}}=0.55\\n_{CHCl_{3}}=0.91\\n_{CCl_{4}}=n_{CH_{4}}-(n_{CH_{3}Cl}+n_{CH_{2}Cl_{2}}+n_{CHCl_{3}})\\n_{CCl_{4}}=0.76mol\\m_{CCl_{4}}=n_{CCl_{4}}*M_{CCl_{4}}\\m_{CCl_{4}}=116.9g[/tex]
To calculate the mass of chlorine we just need to make a mass balance:
[tex]m_{CH_{4}}+m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}\\m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}-m_{CH_{4}}\\m_{Cl_{2}}=243.8g[/tex]
Final answer:
To find the grams of CCl4 formed, we need to calculate the grams of Cl2 reacted with CH4 and then convert it to grams of CCl4 using the stoichiometry of the balanced equation. 2.4 mol of Cl2 reacted with 38.4 g of CH4. Therefore, 369.17 grams of CCl4 were formed.
Explanation:
To find the grams of CCl4 formed, we need to calculate the grams of Cl2 reacted with CH4 and then convert it to grams of CCl4 using the stoichiometry of the balanced equation. We start by calculating the molar mass of CH4, which is 16.04 g/mol. Since all the CH4 reacted, we can use its mass to calculate the moles of CH4. 38.4 g CH4 / 16.04 g/mol = 2.4 mol CH4.
Next, we use the balanced equation to determine the moles of Cl2 that reacted with CH4. From the equation, we know that 1 mole of CH4 reacts with 4 moles of Cl2. So, 2.4 mol CH4 x (4 mol Cl2 / 1 mol CH4) = 9.6 mol Cl2.
Now, we can convert the moles of Cl2 to grams using its molar mass of 70.90 g/mol. 9.6 mol Cl2 x 70.90 g/mol = 681.84 g Cl2 reacted. This is the answer to the second part of the question.
Finally, we use the stoichiometry of the balanced equation to calculate the grams of CCl4 formed. From the equation, we know that 1 mole of CH4 reacts to produce 1 mole of CCl4. So, the moles of CCl4 formed is equal to the moles of CH4 reacted, which is 2.4 mol.
Finally, we convert the moles of CCl4 to grams using its molar mass of 153.82 g/mol. 2.4 mol CCl4 x 153.82 g/mol = 369.17 g CCl4. Therefore, 369.17 grams of CCl4 were formed.
Consider the redox reaction below.
Which equation is a half reaction that describes the reduction that is taking place?
Answer:
1)
Explanation:
Fe+2 + 2e -> Fe
The element that is reduced gains electrons. In this case Fe is gaining electrons
Answer: The half reaction that describes the reduction reaction is [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]
Explanation:
Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)[/tex]
The half reactions follows:
Oxidation half reaction: [tex]Mg(s)\rightarrow Mg^{2+}(aq.)+2e^-[/tex]
Reduction half reaction: [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]
Hence, the half reaction that describes the reduction reaction is [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]
need help with this chem question thanks
Answer:
The answer to your question is letter A.
Explanation:
Isomers are molecules that have the same molecular formula but have a different structure. The molecule from which are looking an isomer has 5 carbons and 1 double bond. Then we need to look for another molecule with these components.
A.- This molecule has 5 carbons and 1 double bond, This structure is an isomer of the first one.
B.- This molecule has 3 carbons and 1 double bond, it's not an isomer of the first structure.
C. This molecule has 4 carbons and 1 triple bonds, it's not an isomer of the first structure.
D. This molecule has 5 carbons but it doesn't have any double bond, then it's not an isomer of the first structure.
A chemical reaction that releases heat to the surroundings is said to be __________ and has a __________ H at constant pressure.a) endothermic, positive b) endothermic, negativec) exothermic, negatived) exothermic, positivee) exothermic, neutral
Answer:
The correct option is c) exothermic, negative.
Explanation:
Reactions that releases heat to the surroundings are called exothermic, and are characterized by negative entalpy (ΔH) values.
A chemical reaction that releases heat to the surroundings is said to be exothermic and has a negative H at constant pressure.
What is an Exothermic reaction?This is the type of reaction which involves the release or loss of heat to the surrounding.
Since heat is lost, exothermic reaction has a negative enthalpy thereby making option C the most appropriate choice.
Read more about Exothermic reaction here https://brainly.com/question/1831525
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Some consequences of ionization in human cells include:
I. creation of unstable atoms.
II. production of free electrons.
III. creation of reactive free radicals capable of producing poisonous substances.
Answer:
number 2 your welcome good luck
Final answer:
Ionization in human cells due to high-energy particles or electromagnetic waves from radioactive nuclides can lead to unstable atoms, free electrons, and reactive free radicals, causing serious disruptions in cell functions and health.
Explanation:
The consequences of ionization in human cells can lead to various forms of radiation damage, which is highly concerning due to the critical functions that cells play in an organism's health. When high-energy particles such as alpha and beta particles, or electromagnetic waves emitted by radioactive nuclides, encounter living cells, they have the potential to cause significant disruption. These disruptions include:
The creation of unstable atoms due to the ionization of molecules within the cells which can cause changes in the atomic structure, rendering the atoms reactive and unstable.The production of free electrons as the radiation breaks chemical bonds, these electrons can then go on to cause further damage within the cell.The creation of reactive free radicals, particularly hydroxyl radicals, which have the potential to disturb physiological processes by damaging biological molecules, such as DNA, proteins, and cell membranes, and can lead to the formation of poisonous substances.The damage caused by ionizing radiation includes both somatic and genetic impacts, with the former affecting the individual's own body cells and the latter having potential consequences for future generations. Cells that reproduce rapidly, such as those found in bone marrow and the gastrointestinal tract, are often most susceptible to this type of damage.
Which is the limiting reagent in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS A) Na2S B) CuSO4 C) Na2SO4 D) CuS E) not enough information
Final answer:
The limiting reagent in this reaction is CuSO4.
Explanation:
The limiting reagent in this reaction is CuSO4.
To determine the limiting reagent, we need to compare the number of moles of each reactant with the stoichiometric ratio of the balanced equation. First, calculate the number of moles of Na2S and CuSO4 using their molar masses. Then, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of the product is the limiting reagent.
In this case, calculate the moles of Na2S and CuSO4. The stoichiometric ratio is 1:1 for Na2S and CuSO4. Since the moles of CuSO4 is smaller than the moles of Na2S, CuSO4 is the limiting reagent.
Nitric acid is usually purchased in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. How much concentrated solution would you take to prepare 1.00 L of 0.120 M HNO3 by mixing with water?
To prepare 1.00 L of 0.120 M HNO3 by mixing with water, you would need approximately 0.992 kg of the concentrated solution.
Explanation:To prepare 1.00 L of 0.120 M HNO3, we can start by calculating the number of moles of HNO3 needed, using the given concentration and volume. We can then use the density of the concentrated solution to calculate the mass of the concentrated solution required. Finally, we can use the percentage of HNO3 in the concentrated solution to determine the mass of HNO3 needed.
Step 1: Calculate the number of moles of HNO3 needed:
moles HNO3 = concentration x volume = 0.120 M x 1.00 L = 0.120 moles HNO3
Step 2: Calculate the mass of the concentrated solution needed:
mass concentrated solution = volume x density = 1.00 L x 1.41 g/mL = 1.41 kg
Step 3: Calculate the mass of HNO3 needed:
mass HNO3 = mass concentrated solution x percentage HNO3 = 1.41 kg x 70.3% = 0.992 kg
Therefore, you would need approximately 0.992 kg of the concentrated solution to prepare 1.00 L of 0.120 M HNO3 by mixing it with water.
What must break in order for water to change from solid to liquid to gas?
a. covalent bonds between water molecules
b. ionic bonds within water molecules
c. hydrogen bonds within water molecules
d. hydrogen bonds between water molecules
e. covalent bonds within water molecules
Hydrogen bonds between water molecules must break for water to change from solid to liquid to gas.
Explanation:In order for water to change from solid to liquid to gas, the hydrogen bonds between water molecules must break. These hydrogen bonds are weaker than covalent or ionic bonds and are constantly forming and breaking in liquid water. When the heat is increased, the higher kinetic energy of the water molecules causes the hydrogen bonds to break completely, allowing water molecules to escape into the air as gas. On the other hand, when water is cooled and freezes, the water molecules form a crystalline structure maintained by hydrogen bonding, making ice less dense than liquid water.
The element rubidium has two naturally occurring isotopes. The atomic mass of 85Rb (72.17 percent abundant) is 84.911794 amu. Determine the atomic mass of 87Rb (27.83 percent abundant). The average atomic mass of Rb is 85.4678 amu.
Answer: 85.47u
Explanation:
The average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.
Now, weighted average simply means that each isotope contributes to the average atomic mass of the element proportionally to its percent abundance.
avg. atomic mass=∑i(isotopei×abundancei)
The more abundant an isotope is, the more its atomic mass will influence the average atomic mass of the element.
In your case, you know that rubidium has two stable isotopes
85Rb → 84.91 u, 72.16% percent abundance
87Rb →86.91 u, 27.84% percent abundance
When you calculate the average atomic mass, make sure that you use decimal abundance, which is simply percent abundance divided by
100.
So, plug in your values to get
avg. atomic mass = 84.91 u × 0.7216 + 86.91 u × 0.2784
avg. atomic mass = 85.4668 u
Rounded to four sig figs, the answer will be
avg. atomic mass =85.47 u